[Guest]

In a game of jackpot, a contestant is faced with three doors. One of the three doors contains the prize, the other door traps the contestant for 1 hour, and the final door traps the contestant for 2 hours before allowing him to contest again. Every time the contestant returns from a 1hour or 2hour lock up, the doors are reshuffled. How long(hours) will it take on average for a person to win the prize.

[soop]

Ok:

We can assume that for the puzzle to be solvable, the doors are shuffled, but the contestant still knows which door they have just been through, leaving a 50% chance of a prize.

However, since 1/3 of contestants will get the prize straight away, they have won the prize in close to 0 seconds.

And over a large enough sample, there will be someone unlucky enough to never get the prize which would make the average at some indeterminable point between infinity and 0.

The only way I can think of to solve it, is if the doors are shuffled, but the prizes stay in the same place, meaning:

1/3 get the prize in 0 seconds

of the remaining 2/3, 1/3 will get the prize after 1 hour, 1/3 after 2 hours and 1/3 after 3 hours.

Edited September 26, 2011 by soop

[Guest]

The first time, there's a 1/3 chance of picking each door, so a (1/3 x 2 + 1/3 x 1) 1 hour average wait if it ended there. The second time, there's (1/9 x 2 + 1/9 x 1 + 1/9 x 2 + 1/9 x 1) an average 2/3 hour wait. If you extend that into a series [(2/3)^(n)] and sum it, you get 3 hours. That should be the total average time, then, maybe.

[Guest]

Given that NO knowledge is transferred because of a truly random shuffle;

It becomes a simple average of (0+1+2)/3 = 1 hour. 1/3 times, 0 wait, 1/3 times 1 hour, 1/3 times 2 hours. Happy hunting, and hope for the best over enough iterations.

Edited September 26, 2011 by bekabeh

[superprismatic]

The answer is 3 hours. Here's why:

Let W0 be the expected wait time if we start by picking the door with the prize. Clearly W0 = 0.

Let W1 and W2 be the expected wait times if we start by picking the door with the 1 hour and

2 hour waits respectively. So, W1 = 1 + (W0/3) + (W1/3) + (W2/3) because we have to wait 1

hour and then we either go to the prize door, the 1-hour door, or the 2-hour door, each with

probability 1/3. Similarly, W2 = 2 + (W0/3) + (W1/3) + (W2/3). Now, since W0 = 0, we have the

two equations, W1 = 1 + (W1/3) + (W2/3) and W2 = 2 + (W1/3) + (W2/3). Solving these two, we get

that W1 = 4 and W2 = 5. Thus the overall expected wait is (W0 + W2 + W3)/3 = (0 + 4 + 5)/3 = 3.

Nice problem, nithinroks! Thanks.

[Guest]

Another way to get the answer is as follows:

If W is the expected wait time, W = P(get the prize door)*0 + P(1 hour wait)*(W+1) + P(2 hour wait)*(W+2)

As P(get the prize door)=P(1 hour wait)=P(2 hour wait)=1/3 we get

W=(W+1)/3 + (W+2)/3

Solving this gives W=3

Edited September 26, 2011 by superprismatic
answers should be in spoilers so others can do the problem without accidentally seeing them

[Guest]

I just used a bit of common sense...

I just used the logic that, after three shuffles, the prize (in an unbiased and fair situation) would have been in all three places once. Hence, if the contestant picked the same door 3 times the longest he would have to wait would be three hours (which means that, as it's not the shortest amount of time, this is the average and quickest possible way to definitely get the prize).

Anyway, not much math for me - just common sense.